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Chapter 8: Problem 46

Evaluate the integrals in Exercises \(39-50\). $$\begin{array}{l} \int \frac{1}{\left(x^{1 / 3}-1\right) \sqrt{x}} d x \\ \text { (Hint: Let } \, x=u^{6} \text { .) } \end{array}$$

Short Answer

Expert verified
The integral evaluates to \( 6x^{1/6} + 3 \ln |x^{1/6}-1| - 3 \ln |x^{1/6}+1| + C \).

Step by step solution

01

Substitution

Using the hint, let \( x = u^6 \). Then we have \( dx = 6u^5 \, du \). Substitute these into the integral.
02

Rewrite the Integral

Substituting \( x = u^6 \) and \( dx = 6u^5 \, du \), the integral becomes: \[\int \frac{1}{(u^2-1)\sqrt{u^6}} 6u^5 \, du \]Simplifying, we know that \( \sqrt{u^6} = u^3 \), leading to: \[6 \int \frac{u^5}{(u^2-1)u^3} \, du = 6 \int \frac{u^2}{u^2-1} \, du\]
03

Simplify the Integral

The integral becomes: \[6 \int \frac{u^2}{u^2-1} \, du = 6 \int \left( 1 + \frac{1}{u^2-1} \right) \, du\]This splits into two simpler integrals: \[6 \left( \int 1 \, du + \int \frac{1}{u^2-1} \, du \right)\]
04

Evaluate the Integral

First integral: \[6 \int 1 \, du = 6u\]Second integral can be rewritten using partial fraction decomposition as: \[\frac{1}{u^2-1} = \frac{1}{2} \left( \frac{1}{u-1} - \frac{1}{u+1} \right)\]This results in: \[6 \int \left( \frac{1}{2} \frac{1}{u-1} - \frac{1}{2} \frac{1}{u+1} \right) du\]
05

Solve Remaining Integrals

Integrate each term:\[6 \int \frac{1}{2} \frac{1}{u-1} \, du = 3 \ln |u-1|\]\[6 \int -\frac{1}{2} \frac{1}{u+1} \, du = -3 \ln |u+1|\]Combine results:\[6u + 3 \ln |u-1| - 3 \ln |u+1| + C\]
06

Back-Substitute for x

Replace \( u \) with \( x^{1/6} \) to return to terms of \( x \):\[6x^{1/6} + 3 \ln |x^{1/6}-1| - 3 \ln |x^{1/6}+1| + C\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique in integral calculus that simplifies complex integrals by changing the variable of integration. In this method, we select a new variable, often denoted as \( u \), and express the original variable \( x \) in terms of this new one. This often transforms a difficult integral into one that is easier to handle. The key steps are:
  • Choose an appropriate substitution. In our exercise, we let \( x = u^6 \), which helps in simplifying the expression inside the integral.
  • Find the differential \( dx \) in terms of \( du \). Here, \( dx = 6u^5 \, du \) to account for the change in variables.
  • Substitute \( x \) and \( dx \) in the integral, transforming it into a function of \( u \).
  • Simplify the resulting integral, if needed, and perform the integration.
After integrating with respect to \( u \), don't forget to substitute back the original variable to express the solution in the initial terms.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This makes integration manageable when dealing with rational functions.
  • The goal is to express a complicated fraction like \( \frac{1}{u^2-1} \) as a sum of simpler fractions. In our solution, this expression decomposes into \( \frac{1}{2} \left( \frac{1}{u-1} - \frac{1}{u+1} \right) \).
  • To perform partial fraction decomposition, identify roots of the denominator and express each term as a fraction with those roots. For \( u^2-1 \), the roots are \( u-1 \) and \( u+1 \).
  • The decomposition allows us to integrate each term separately, which is simpler than integrating the initial fraction directly.
This technique highlights the power of algebraic manipulation in integrals, enabling us to integrate each part efficiently.
Definite Integrals
Definite integrals help compute the area under a curve between specified bounds, providing a concrete numerical value instead of an indefinite form. Although this exercise focuses on solving an indefinite integral, understanding definite integrals is crucial when these equations have limits.
  • In a definite integral setting, the expression \( \int_{a}^{b} f(x) \, dx \) is evaluated from \( x = a \) to \( x = b \). This computes the exact area under the curve of \( f(x) \) from \( a \) to \( b \).
  • Transformations of variables, as we've done with substitution, can still apply. It's essential to adjust limits accordingly in definite calculations.
  • Simplifying the integrand, possibly through methods like substitution or partial fractions, remains important since it ensures computations are manageable.
While indefinite integrals give us a general form of antiderivatives, definite integrals provide specific measures of quantities, crucial in applications spanning physics, engineering, and beyond.

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Most popular questions from this chapter

a. Use a CAS to evaluate $$\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x$$ where \(n\) is an arbitrary positive integer. Does your CAS find the result? b. In succession, find the integral when \(n=1,2,3,5,\) and 7 Comment on the complexity of the results. c. Now substitute \(x=(\pi / 2)-u\) and add the new and old integrals. What is the value of $$\int_{0}^{\pi / 2} \frac{\sin ^{n} x}{\sin ^{n} x+\cos ^{n} x} d x ?$$ This exercise illustrates how a little mathematical ingenuity solves a problem not immediately amenable to solution by a CAS.

In Exercises \(21-32,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{\theta^{4}-4 \theta^{3}+2 \theta^{2}-3 \theta+1}{\left(\theta^{2}+1\right)^{3}} d \theta$$

Use a CAS to perform the integrations. Evaluate the integrals a. \(\int x \ln x \, d x \quad\) b. \, \(\int x^{2} \ln x d x \quad\) c. \(\int x^{3} \ln x d x\) d. What pattern do you see? Predict the formula for \(\int x^{4} \ln x d x\) and then see if you are correct by evaluating it with a CAS. e. What is the formula for \(\int x^{n} \ln x d x, n \geq 1 ?\) Check your answer using a CAS.

Require the use of various trigonometric identities before you evaluate the integrals. $$\int \sin ^{2} \theta \cos 3 \theta d \theta$$

Evaluate \(\int_{3}^{\infty} \frac{d x}{x \sqrt{x^{2}-9}}\).
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